"""
3元一张彩票，奖金为1、2、3、4等概率，
连买N张，问不亏本的概率。N<=30
所有可能的奖金的组合为4^N。
求出和大于等于3N的方案总数即可
令 Dij表示买第i张彩票，奖金为j元的方案总数
D[i][j] = D[i-1][j-1] + D[i-1][j-2] + D[i-1][j-3] + D[i-1][j-4]
"""
from math import gcd

N = int(input())
D = [[0 for j in range(N+N+N+N+1)] for i in range(N + 1)]

D[0][0] = 1
for i in range(1, N + 1):
    for j in range(1, N+N+N+N+1):
        D[i][j] += D[i-1][j-1]
        if j >= 2: D[i][j] += D[i-1][j-2]
        if j >= 3: D[i][j] += D[i-1][j-3]
        if j >= 4: D[i][j] += D[i-1][j-4]
ans = 0
for i in range(N+N+N, N+N+N+N+1):
    ans += D[N][i]
fenmu = 4**N
g = gcd(ans, 4**N)
ans //= g
fenmu //= g
print(str(ans)+'/'+str(fenmu))